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The best is to rewrite your transfer function in the proper low-entropy way, with a leading term. I have assumed the roots are all negative leading to left-half-plane pole and zero: $G(s)=\frac{s+5447}{s+961.3}=G_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}$ where: $G_0=\frac{5447}{961.3}=5.66$ $\omega_z=5447$ rds/s or 867 Hz $\omega_p=961.3\... 1 I don't recognize your transfer function as one of the classic low-pass-filter implementations. A Butterworth filter, for instance, would be:  H(s) = \frac{f_c^2}{s^2+1.414f_cs+f_c^2}  note: for me, it's unusual to see the use of$f$instead of$w$in these formulas, but mathematically it should be equivalent, as long as you use$s=jf$instead of ... 0 Are there any obvious problems with the circuit, that may lead to the observed drifting behavior? The minimum supply for an LPV821 is 1.7 volts and you are expecting it to work above 1.6 volts. Your AC current is 1 uA and this produces a voltage (across the 20 ohm resistor) of maybe 28 uVp-p. Given that your op-amp is running without feedback AND it has ... 0 These two questions are nearly unrelated to each other. In normal situation the internal ICs usually are powered with 3.3V, so typically a 5V-to-3.3V regulator is required. And therefore the bypass caps are de-coupled from the 5-V input. Now, each IC needs de-coupling caps, and their values are usually specified in respective datasheets, with all this ... 1 Can you reference(or link) the component datasheets to see what the manufacturer recommends for decoupling capacitors? It is likely that the components dont really need their own 10uF cap. Often times just a 0.1uF or a 0.1uf + 1-4.7uF will be recommended. I would expect that an individual 0.1uF plus a shared 4.7 or 10uF would work just fine. 1 Do read my answer in Output capacitors placement in PCB Is using only 1 piece of 10uF at the end of filter instead of placing 3 pieces 10uF for each ic enough? Yes, 1 capacitor of 10uF will serve all other IC's. I wonder if you really need 10uF and think you could use 2.2uF or 4.7uF instead. But that depends on cable length and whether the input current ... 0 Dwayne is correct. Because you want a very low frequency filter, the resistors and capacitors have to be very large. All real op amps have currents that flow into or out of their input terminals. The large resistors cause this current to create a DC offset on the op amps. You want to use an op amp with a CMOS, or perhaps JFET input stage, so the input ... 0 Can L-filter be ever used (so easily) to filter power, and what are the best practices? I pretty much always (when implementing an LC low pass filter to remove noise on power rails) design it as an RC LP filter and put the inductor in parallel with the resistance so that DC currents are supported by the very low impedance of the inductor (as opposed to ... 1 Well, we have the following circuit: simulate this circuit – Schematic created using CircuitLab Using KCL, we can write:  \begin{cases} \text{I}_{\text{R}_1}+\text{I}_{\text{C}_2}=\text{I}_{\text{R}_2}\\ \\ \text{I}_{\text{R}_2}=\text{I}_{\text{C}_1} \end{cases}\tag1  Using KVL, we can write:  \begin{cases} \text{I}_{\text{R}_1}=\frac{\text{... 2 Your transfer function$ \frac {V_{out} }{ V_{in} } \\$ is correct. However, the error is in the following: $$H(s)=1+\frac{R_{1}}{R_{2}} \cdot \frac{1}{1+R_{1}Cs} \neq \bigg(1+\frac{R_{1}}{R_{2}} \bigg)\frac{1}{1+R_{1}Cs}$$ You are also right that the DC gain is $$H(0)=1+\frac{R_{1}}{R_{2}} \cdot \frac{1}{1+R_{1}Cs} = 1+\frac{R_{1}}{R_{2}}$$ There is ...

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The transfer function of the non-inverting circuit (which is no lowpass) is not correct. There is a simple math error. The correct expression is: Vout/Vin=1+ (R1/R2)*[1/(1+sR1C1) Hence, for s approaching infinity, the transfer function is approaching "1" (and not zero).

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It really depends on what you want to protect: The input power from the buck converter, or whatever is being powered by the buck converter. A buck converter primarily causes ripple and noise on its input, due to the fast rise times of most switches used in such converters. You generally place as much quality, ceramic capacitance as you are willing to ...

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